# Problem in understanding the relation between relative error and the number of correct significant figures

Here is a theorem from my textbook relating the relative error with number of correct significant figures:

Theorem: If the first significant figure of a number is $$k$$ and the number is correct to $$n$$ significant figures, then the relative error is less than $$1/\left(k \times 10^{n-1}\right)\;.$$

Proof: Let $$N$$ be the exact value of the number, $$n$$ be the number of significant figures, $$m$$ be the number of correct decimal places.

Three cases must be distinguished, namely $$(a) ~m\lt n, ~(b)~m= n$$ and $$(c) ~m\gt n\;.$$

## Case I:

Here the number of digits in the integral part of $$N$$ is $$n-m\;.$$ Denoting the first significant digit of $$N$$ by $$k,$$ we have \begin{align}E_\textrm{a} &\equiv \textrm{absolute-error} \leq \frac12 \cdot 10^{-m},\\ N & \geq \color{red}{k\cdot 10^{n-m-1}-\frac12 \cdot 10^{-m}} \;.\end{align}

Hence \begin{align}E_\textrm r &\equiv \textrm{relative-error}\leq \frac{\frac{1}{2}\cdot 10^{-m}}{\color{red}{k\cdot 10^{n-m-1}-\frac12 \cdot 10^{-m}}} = \ldots \end{align}

## Case II:

$$N$$ is a pure fraction and $$k$$ is the first digit after the decimal point.

Then \begin{align}E_\textrm r& \leq \frac{\frac{1}{2}\cdot 10^{-m}}{\color{red}{k\cdot 10^{-1}-\frac12 \cdot 10^{-m}}} = \ldots \end{align}

## Case III:

In this case, $$k$$ occupies the $$(m-n+1)$$th decimal place and therefore \begin{align}E_\textrm r& \leq \frac{\frac{1}{2}\cdot 10^{-m}}{\color{red}{k\cdot 10^{n-m-1}-\frac12 \cdot 10^{-m}}} = \ldots \end{align}

[...]

But I cannot get how the author deduces $$N \geq k\cdot 10^{n-m-1}-\frac12 \cdot 10^{-m}\;.$$

Please anyone help me clear this doubt. Thanks in advance.

## 1 Answer

Let's try an example: $$N \approx 4034.56,$$ where all the digits are known to be correct.

In this example, $$n = 6,$$ $$m = 2,$$ and $$k = 4.$$ So $$n - m - 1 = 3$$ and $$k \cdot 10^{n - m - 1} = 4 \cdot 10^3 = 4000.$$ Also, $$\frac12 \cdot 10^{-m} = \frac12 \cdot 10^{-2} = 0.005.$$ So

$$k\cdot 10^{n-m-1}-\frac12 \cdot 10^{-m} = 4000 - 0.005 = 3999.995,$$

and it is clear that $$N$$ is greater than that.

The basic idea here is that $$k \cdot 10^{n - m - 1}$$ is the place value of the most significant digit. The actual value $$N$$ might be larger, but it can't be much smaller than that. The $$-\frac12 \cdot 10^{-m}$$ part represents how much smaller $$N$$ can possibly be.

More generally, the worst case we have to worry about is where all the known correct digits after the first significant digit are zero. For example, if the number were written $$4000.00,$$ the actual value of $$N$$ might be $$3999.996.$$ But it is not $$3999.994.$$ Anything less than $$3999.995$$ would be written with fewer than four digits left of the decimal point or with a leading digit less than $$4.$$

By the way, if the known correct digits were written in scientific notation, there would only need to be one case to examine, not three. But that would be a different proof.